# Coclosed form is sum of coexact and harmonic form.

Let $$(\mathcal{M},g)$$ be a compact and connected Riemannian manifold, $$\mathrm{d}$$ and $$\delta$$ differential and codifferential, respectively, and $$\Delta:=\delta\mathrm{d}+\mathrm{d}\delta$$ the corresponding de Rham-Hodge Laplacian. Is it true that

$$\mathrm{ker}(\delta)=\mathrm{ran}(\delta)\oplus\mathrm{ker}(\Delta)$$

Direction "$$\supset$$" is clear, since $$\delta^{2}=0$$ and every harmonic form is both closed and coclosed. However, I am struggeling with the other direction. Let $$\omega\in\mathrm{ker}(\delta)$$. My idea was to use condradiction: Assume $$\omega\notin\mathrm{ran}(\delta)\oplus\mathrm{ker}(\Delta)$$. Is it possible to argue by Hodge decomposition that then $$\omega\in\mathrm{ran}(\mathrm{d})$$? Because then the claim follows, since $$\omega$$ would be both closed and coclosed and hence in particular harmonic.

EDIT: At least for $$1$$-forms, it should be true: Take $$A\in\Omega^{1}(\mathcal{M})$$ such that $$\delta A=0$$. By Hodge decomposition, $$A=\mathrm{d}f+\delta F+h$$ for $$h$$ harmonic. Then $$\delta A=\Delta f=0$$ and hence $$f=\mathrm{const}$$, since any harmonic $$1$$-form on a compact manifold is necessarily constant. It follows that $$A=\delta F+h\in\mathrm{ran}(\delta)\oplus\mathrm{ker}(\Delta)$$.

You can use a similar argument as in your EDIT: Take $$\omega\in\Omega^{k}(\mathcal{M})$$ such that $$\delta\omega=0$$. By Hodge decomposition,

$$\omega=\mathrm{d}\alpha+\delta\beta+h$$

for $$\alpha\in\Omega^{k-1}(\mathcal{M})$$, $$\beta\in\Omega^{k+1}(\mathcal{M})$$ and $$h\in\mathcal{H}^{k}(\mathcal{M})$$. It follows that

$$\delta\omega=\delta\mathrm{d}\alpha\stackrel{!}{=}0$$

$$\delta\mathrm{d}$$ is for $$k\neq 0$$ different from $$\Delta$$. However, since also $$\mathrm{d}^{2}\alpha=0$$, you see that $$\mathrm{d}\alpha$$ is both closed and coclosed and hence in particular harmonic, i.e. $$\mathrm{d}\alpha\in\mathcal{H}^{k}(\mathcal{M})$$. This shows that

$$\omega=\delta\beta+(h+\mathrm{d}\alpha)$$

gives the required decomposition.

Since $$\delta$$ is the formal adjoint of $$d$$, for any form $$\alpha$$ satisfying $$\delta d \alpha = 0$$ we have $$\langle d\alpha, d\alpha \rangle = \langle \delta d \alpha, \alpha\rangle = 0,$$ which implies $$d \alpha = 0.$$ So, your argument for $$1$$-forms holds for a general form.